How do you factor $x^{6} + 2 x^{5} + 3 x^{4} + 6 x^{3}$ $x^6 + 2x^5 + 3x^4 + 6x^3$ ?

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How do you factor $x^{6} + 2 x^{5} + 3 x^{4} + 6 x^{3}$ $x^6 + 2x^5 + 3x^4 + 6x^3$ ?

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Answer 1 · 2015-11-15T19:40:27

Separate out the common factor $x^{3}$ $x^3$ then factor by grouping to find:

$x^{6} + 2 x^{5} + 3 x^{4} + 6 x^{3} = x^{3} (x^{2} + 3) (x + 2)$ $x^6+2x^5+3x^4+6x^3 = x^3(x^2+3)(x+2)$

Explanation:

$x^{6} + 2 x^{5} + 3 x^{4} + 6 x^{3}$ $x^6+2x^5+3x^4+6x^3$

$= x^{3} (x^{3} + 2 x^{2} + 3 x + 6)$ $=x^3(x^3+2x^2+3x+6)$

$= x^{3} ((x^{3} + 2 x^{2}) + (3 x + 6))$ $=x^3((x^3+2x^2)+(3x+6))$

$= x^{3} (x^{2} (x + 2) + 3 (x + 2))$ $=x^3(x^2(x+2)+3(x+2))$

$= x^{3} (x^{2} + 3) (x + 2)$ $= x^3(x^2+3)(x+2)$

This has no simpler factors with Real coefficients since $x^{2} + 3 \geq 3 > 0$ $x^2+3 >= 3 > 0$ for all $x in RR$

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How do you factor $x^{6} + 2 x^{5} + 3 x^{4} + 6 x^{3}$ $x^6 + 2x^5 + 3x^4 + 6x^3$ ?

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